JEE Advanced 2024 · Paper 1 · Mathematics Q16

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MathematicsList match+3 / -1Question 16 of 51

Let $\gamma \in \mathbb{R}$ be such that the lines $L_1 : \dfrac{x+11}{1} = \dfrac{y+21}{2} = \dfrac{z+29}{3}$ and $L_2 : \dfrac{x+16}{3} = \dfrac{y+11}{2} = \dfrac{z+4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$ . Let $O = (0, 0, 0)$ , and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$ .

Match each entry in List-I to the correct entry in List-II.

The correct option is

List-I

P
$\gamma$ equals
Q
A possible choice for $\hat{n}$ is
R
$\overrightarrow{OR_1}$ equals
S
A possible value of $\overrightarrow{OR_1}\cdot\hat{n}$ is

List-II

1
$-\hat{i} - \hat{j} + \hat{k}$
2
$\sqrt{\dfrac{3}{2}}$
3
$1$
4
$\dfrac{1}{\sqrt{6}}\hat{i} - \dfrac{2}{\sqrt{6}}\hat{j} + \dfrac{1}{\sqrt{6}}\hat{k}$
5
$\sqrt{\dfrac{2}{3}}$

P → 3; Q → 4; R → 1; S → 2

P → 5; Q → 4; R → 1; S → 2

P → 3; Q → 4; R → 1; S → 5

P → 3; Q → 1; R → 4; S → 5

Answer and worked solution

Official answer: C

Intersection. $L_1$ : $(-11+s,\ -21+2s,\ -29+3s)$ ; $L_2$ : $(-16+3t,\ -11+2t,\ -4+\gamma t)$ .

From $x$ : $s = 3t - 5$ .
From $y$ : $-21 + 6t - 10 = -11 + 2t \implies t = 5$ , so $s = 10$ .

$R_1 = (-1,-1,1)$ . From $z$ : $1 = -4 + 5\gamma \implies \gamma = 1$ . (P)→(3).

Normal. $(1,2,3)\times(3,2,1) = (-4, 8, -4) \parallel (1,-2,1)$ , so $\hat n = \frac{1}{\sqrt6}(1,-2,1)$ . (Q)→(4).

$\overrightarrow{OR_1} = -\hat i - \hat j + \hat k$ . (R)→(1).

$\overrightarrow{OR_1}\cdot\hat n = \dfrac{-1+2+1}{\sqrt6} = \sqrt{\tfrac23}$ . (S)→(5).

Option (C).

Topics:Equation of a line in space Cross product Equation of a plane

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JEE Advanced 2024 · Paper 1, Mathematics question 16